The standard definition is that a function $f:\mathbb{R}^n\to \mathbb{R}$ is *differentiable* at a point $x$ if there exists a linear map $\mathrm{d}f_x: \mathbb{R}^n \to \mathbb{R}$ such that

$$f(x+h) = f(x) + \mathrm{d}f_x(h) + \epsilon \|h\|$$

where $\epsilon\to 0$ as $h\to 0$. This is stronger than the existence of all partial (or directional) derivatives, but weaker than their continuity. However, when talking about higher differentiability, one usually switches to talking about partial derivatives, asking them to be continuous in order to prove basic properties.

Suppose that instead we define $f$ to be "twice differentiable" at $x$ if in addition to $\mathrm{d}f_x$ as above there exists a quadratic form $\mathrm{d}^2f_x$ such that

$$f(x+h) = f(x) + \mathrm{d}f_x(h) + \frac{1}{2}\mathrm{d}^2f_x(h) + \epsilon \|h\|^2$$

where $\epsilon\to 0$ as $h\to 0$. This is true if $f$ has continous second-order partials (it's the multidimensional Taylor expansion, with $\mathrm{d}^2f_x$ the Hessian matrix).

- Does this imply that all second-order partial derivatives of $f$ exist?
- If so, does it imply that the mixed second-order partials are equal?

exactlylike the 1-dimensional case (with the mess of factorials hidden away within a clean formalism); of course, one can bust out coordinates and recover the usual messier explicit version from that. $\endgroup$ – user76758 May 6 '14 at 21:47The Way of Analysis, after proving the second-order Taylor approximation to a $C^2$ function, he says "We will not discuss the problem of obtaining a converse kind of statement, deducing the existence of the second derivative from the existence of quadratic polynomial approximations, because such theorems are extremely difficult to prove and have few applications." But that implies that such theoremsexist, which the answers to this question seem to deny. Can anyone guess what he had in mind? $\endgroup$ – Mike Shulman May 12 '14 at 22:1611more comments